Along the lines of what Ray Yang suggested, for $f \in L^p(\Omega)$ with $\Omega \subset \mathbb{R}^n$ and $p > n$ the gradient in fact has a Holder $1-n/p$ modulus of continuity. One obtains this either by estimating the Newtonian potential or by applying $W^{2,p}$ estimates and Morrey's embedding.
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Connor MooneyJul 10 '13 at 12:40

2 Answers
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Of course the regularity theory is local, then is is the same on a smooth manifold than in an open set of $\mathbb{R}^n$. Then the answer of your question is found in Sobolev embedding and regularity theory. If $f\in L^p$ with $p>1$ then $u\in W^{2,p}$ , see chapter 9 of Gilbarg&Trudinger. Then $W^{2,p}\subset W^{1,\infty}$ if $p> n$.
But using Lorentz space, see Grafakos's book , you can improved a bit this result. $f\in L^{n,1}$ implies $u \in W^{2,(n,1)}$, then with the improved Sobolev embedding: $W^{1,(n,1)}\subset L^{\infty}$, we get $\nabla u\in L^\infty$.